I often see identity
$$\sum_{k=0}^{n-1}e^{\tau ika/n} = \cases {n \quad \text{ if }n | a\\0\quad \text{ otherwise}}$$
in the context of generating functions. It allows to zero out all members of sequence, indexed by $a$, except every $n$-th one (n divides a). The textbooks seem to consider it so obvious that it does not need any proof. Is it right or there is one?
On
The sum becomes zero when multiplied by $\exp(\tau\mathbf ia/n)-1$, which factor is nonzero when $n\not\,\mathrel\backslash a$, so the sum must have been $0$ already in this case. But when $n\mathrel\backslash a$ then $\exp(\tau\mathbf ika/n)=1^k=1$ and the sum becomes $\sum_{i=0}^{n-1}1=n$.
On
The intuitive geometric approach is to look at all the $n$th roots. I mean, draw them as points on a piece of paper (or in your head). See how nicely they're distributed, all evenly along the unit circle. That means their sum is equal to $0$.
Now raise them to some power. If that power is a multiple of $n$, then they will all end up at $1$, and their sum will be $n$. However, if the power is not a multiple of $n$, they will still be spread out. They might end up some on top of others, in which case you will have "piles" of points, where each pile has $\gcd(n, a)$ points. So each pile will weigh the same as any other pile, and they're still evenly spread out along the unit circle (although possibly a bit sparser). Therefore their sum is $0$.
On
The numbers $x_k =e^{2\pi i \frac{k}{n}}$, $0 \le k \le n-1$ are the roots of the polynomial $x^n-1$. The sums $S_a= \sum_{k=0}^{n-1} x_k^a$ have period $n$ in $a$ ( $S_{a+n} = S_a$ ) so it is enough to prove the assertion for $0\le a \le n-1$. For $a=0$ it's true (easy check). Let's show that $S_1, \ldots, S_{n-1} =0$. We can show in fact that for every symmetric polynomial in $x_k$ homogenous of degree $d$, $1\le d \le n-1$ we have $P(x_1, \ldots, x_n) =0$. Indeed, $P(x_1,\ldots, x_k)$ will be expressed in terms of the fundamental symmetric polynomials in $x_k$, $e_1 = \sum x_k$, $e_2 = \sum x_k x_l$, $e_n = x_1 \ldots x_n$. Now from Viete relations we have $e_1 = ...= e_{n-1} = 0$. Now any monomial in the expression of $P(x_1, \ldots, x_n)$ in terms of the $e_i$ involves only $e_1$, $\ldots e_{n-1}$ and is of degree $>0$. Therefore, every such monomial evaluated in $e_1$, $\ldots e_n$ give $0$, and so $P(x_1, \ldots, x_n)= 0$.
On
For $n$, $a$ integers, $n>0$ the map
$$ k\!\!\!\!\mod n \mapsto a k \!\!\!\!\mod n$$ maps $\mathbb{Z}/n $ onto $\gcd (a,n)\cdot \mathbb{Z}/n\simeq \mathbb{Z}/(n/d)$ with fibres of cardinality $\gcd (a,n)=\colon d$. Therefore we have
$$S_a = \gcd (a,n)\cdot S'_1$$
where the sum $S'$ is the sum of the roots of the equation $X^{\frac{n}{d}}-1$, and this from Viete's equalities is $0$ is $d < n$, and $n$ if $d= n$.
(Assuming that $\tau$ in your formula stands for $2 \pi$) this is an immediate consequence of the "well-known" formula for the (finite) geometric sum $$ \sum_{k=0}^{n-1} x^k = \cases {n \quad \quad \text{ if } x = 1\\ \frac{x^n-1}{x-1}\quad \text{ otherwise}} $$ applied to $x = e^{\tau ia/n}$. (Note that $x = 1$ exactly if $n$ divides $a$.)